



 
Hi Brandon, I drew a diagram of what I see from your description. Since the angle $BDA$ is a right angle \[\tan(16^o) = \frac{x}{2} \mbox{ and hence } x = 2 \times \tan(16^o) = 2 \times 0.2867 = 0.5735.\] Hence the diameter of the circular base of my truncated cone is $0.19 + 2 \times 0.5735 = 1.34$ inches. I hope this helps, 



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